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Let $\Omega = \{1,\dots, n\}$, with $n \ge 2$, $\mathcal{A}=\mathcal{P}(\Omega)$, and $P$ is determined thus: $\epsilon$ is an irrational number in $(0,(n-1)^{-1})$, $P({k}):=\epsilon$ for $2\le k\le n$, $P({1}):=1-(n-1)\epsilon$.

Show that $\{\emptyset, \Omega\}$ is the only pair of independent events in $\mathcal{A}$, i.e., this space is independence-free.

I need to show that any finite collection of subsets of $\{1,\dots, n\}$ is not independent. But I don't know how to formulate an arbitrary collection to show this. I would greatly appreciate any hints or solutions.

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You don't have to formulate an arbitrary collection of sets; it suffices to show that no two sets in $\mathcal A$ are independent. This is because te existence of an independent collection of size more than two implies the existence of an independent pair. If you wanted to talk about an arbitrary (finite) collection of events, you would start by saying "Let $E_1,E_2,\dots,E_k$ be elements of $\mathcal A$ . . .", then proceed to prove $P(E_1)\cdots P(E_k)\neq P(E_1\cap \cdots \cap E_k)$.

Let $E,F$ be events in $\mathcal A$. Let $p,q$ and $r$ be the number of elements of $E, F$ and $E\cap F$, respectively. The goal is to compute $P(E)P(F)$ and $P(E\cap F)$, then prove they are unequal. These probabilities depend on whether $1\in E$ and whether $1\in F$, so you need to split into cases.

  1. $1\notin E,1\notin F$. In this case, $$ P(E)P(F)= (p\epsilon)(q\epsilon)=pq\epsilon^2,\qquad P(E\cap F)=r\epsilon $$ The goal is then to show $pq\epsilon^2\neq r\epsilon $. If the equation were true, it would imply $\epsilon = r/pq$, contradicting its irrationality.

  2. $1\in E, 1\notin F$. $$ P(E)P(F)= ((p-1)\epsilon\cdot(1-(n-1)\epsilon))(q\epsilon),\qquad P(E\cap F)=r\epsilon $$ Equating, and rearranging, $$ 0 = (n-1)\epsilon^2-\epsilon +r/(q(p-1)) $$ Using the quadratic formula, you get a nasty formula for $\epsilon$ which involves $\sqrt{1-4(n-1)r/(q(p-1))}$. In case $r\ge1$, you can show the argument of the square root is negative, so no solutions exist (hint: $(p-1)+q=(n-1)$. Maximize the product $xy$ contrained to $x+y=$ constant). In case $r=0$, the events $E,F$ are disjoint, so certainly not independent except in the trivial cases where they are both empty or full.

  3. $1\in E\cap F$. The messiest case, left to reader.